In an isothermal reversible expansion, if the volume of 96 gm of oxygen at $27^{\circ} C$ is increased from 70 litres to 140 litres, then the work done by the gas will be

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Solution

The correct option is D
$W = μ R T l o g_{e} \frac{V_{2}}{V_{1}}$
$= (\frac{m}{M}) R T l o g_{e} \frac{V_{2}}{V_{1}} = 2.3 \times \frac{m}{M} R T l o g_{10} \frac{V_{2}}{V_{1}}$
$= 2.3 \times \frac{96}{32} R (273 + 27) l o g_{10} \frac{140}{70} = 2.3 \times 900 R l o g_{10} 2$

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In an isothermal reversible expansion, if the volume of 96 gm of oxygen at 27∘C is increased from 70 litres to 140 litres, then the work done by the gas will be

The correct option is D W=μRT logeV2V1 =(mM)RT logeV2V1=2.3×mMRT log10V2V1 =2.3×9632R(273+27)log1014070=2.3×900R log102

In an isothermal reversible expansion, if the volume of 96 gm of oxygen at $27^{\circ} C$ is increased from 70 litres to 140 litres, then the work done by the gas will be

The correct option is D
$W = μ R T l o g_{e} \frac{V_{2}}{V_{1}}$
$= (\frac{m}{M}) R T l o g_{e} \frac{V_{2}}{V_{1}} = 2.3 \times \frac{m}{M} R T l o g_{10} \frac{V_{2}}{V_{1}}$
$= 2.3 \times \frac{96}{32} R (273 + 27) l o g_{10} \frac{140}{70} = 2.3 \times 900 R l o g_{10} 2$